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2 years ago

When a warm front passes your location, the temperature rises rapidly. As a result, air density:

Chemistry

2 answers:

sukhopar [10]2 years ago

8 0

The air density decreases.

deff fn [24]2 years ago

3 0

Answer:

decreases

Explanation:

As you know when the temperature of a medium decreases, the density increases, the majority of the solids present in nature depende of a certain temperature, when they heat up they tend to go to a liquid form and then to gas, so if the temperature in the are rises rapidly the densitiy of the air would decrease as it is expanding because of the rise of the temperature.

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You will be climbing a mountain with very high elevation. What instrument should you bring along to determine the atmospheric pr

slamgirl [31]

1) b
2)c
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4)c

8 0

3 years ago

Read 2 more answers

Which is the correctly balanced chemical equation for the reaction of KOH and H2SO4?

zalisa [80]

Answer:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O.

Explanation:

This is an acid-base reaction which produces salt and water according to the balanced equation:

it is clear that 2.0 moles of KOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of K₂SO₄ and 2.0 moles of H₂O

4 0

2 years ago

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction = a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0

2 years ago

How many valence electrons does an atom of al possess?

Irina18 [472]

Answer:

Explanation:

Third element to right

3 0

2 years ago

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

2 years ago